significance.Rmd

---
title: "Biological Significance"
author: "Tina Lasisi"
date: "`r Sys.time()`"
output:
  workflowr::wflow_html:
    toc: yes
    number_sections: yes
  pdf_document: default
editor_options:
  chunk_output_type: console
---

```{r setup, message=FALSE, warning=FALSE, include=FALSE, paged.print=FALSE}

library(tidyverse)
library(knitr)
library(cowplot)

F = rprojroot::is_rstudio_project$make_fix_file()

knitr::opts_chunk$set(echo = FALSE, include = TRUE, eval = TRUE, warning = FALSE, message = FALSE, fig.retina = 2, fig.width = 8, fig.height = 4, out.width = "100%")

```


```{r functions, include=FALSE}
plot_path = F("output/")

pltsave_func <- function(plot, plot_path, width, height){
  ggsave(
      filename = paste0(deparse(substitute(plot)), ".png"),
      plot = plot, 
      path = plot_path,
      width = width,
      height = height)
  plot(plot)
}

```


```{r read_df_admixed, include=FALSE}

df_admixed <- read_csv(F("data/df_Admixed_Merged.csv"))
colnames(df_admixed)[c(18:22)] = c("SAS","AMR","EUR","AFR","EAS")


```

To explore the significance of quantifying hair fiber morphology, we explore the relationship between various quantitative hair traits, categorical data and genotype data on the same sample. 


```{r df_admixed_plt, include=FALSE}

df_admixed_plt <- df_admixed %>%
  group_by(ID, hair_texture, HairTypeBin, sex, age) %>%
  summarize(area_mean = mean(area),
         area_median = median(area),
         min_mean = mean(min),
         max_mean = mean(max),
         min_median = median(min),
         max_median = median(max),
         eccentricity_mean = mean(eccentricity),
         eccentricity_median = median(eccentricity),
         curv_median = mean(curv_median),
         curv_mean = mean(curv_mean),
         length_mean = mean(length_mean),
         length_median = mean(length_median),
         SAS = mean(SAS),
         AMR = mean(AMR),
         EUR = mean(EUR),
         AFR = mean(AFR),
         EAS = mean(EAS),
         hair_count = mean(hair_count),
         m_index = mean(m_index))  %>%
  mutate(hair_texture = str_to_title(hair_texture)) %>% 
  filter(hair_count > 4 & AFR > 0.1)

df_admixed_plt$hair_texture <- factor(df_admixed_plt$hair_texture, 
                                     levels = c("Straight","Wavy",
                                                "Curly",
                                                "Very Curly",
                                                NA))

df_admixed_plt$HairTypeBin <- factor(df_admixed_plt$HairTypeBin, 
                                     levels = c("Straight","Wavy",
                                                "Curly",
                                                "Very Curly"))

# I filtered out anyone who had less than 5 hair fragments and I only kept the "first" reported ethnicity to make sure I didn't have duplicates of IDs

```

Our data consists of 193 individuals for whom we have quantitative hair phenotype data. In our first data quality control step, we filter to keep individuals who have more than 4 hair fragments in their curvature image and over 10% African ancestry. We calculate mean and median values for the cross-sectional data we have collected for individuals (~ 6 sectioned hair fibers). In our analyses, we use median values as they are less affected by intra-individual outliers.


# Self-reported hair texture vs. quantitative hair curvature

We compare the self-reported hair texture with mean and median curvature for our sample. 

```{r plt_admixed_SelfRepHair, fig.height=4, fig.width=8, fig.cap= "Self-reported hair texture vs. quantitative hair curvature"}


plt_admixed_SelfRepHair_CurvMean <- df_admixed_plt %>%
  filter(is.na(hair_texture)=="FALSE")%>%
  ggplot(., aes(hair_texture,
                      curv_mean))+
  geom_violin(fill="#a6bddb",
              scale = "width")+
  geom_point(position = position_jitter(width=0.1),
             alpha = 0.4)+
  theme(axis.text.x = element_text(angle=90))+
  labs(x = "Self reported hair texture",
       y = bquote("Mean curvature ("*mm^-1*")"),
       title = "Quantitative variation underlying \n self reported hair texture (mean)")+
  theme_bw() +
  coord_flip()


plt_admixed_SelfRepHair_CurvMedian <- df_admixed_plt %>%
  filter(is.na(hair_texture)=="FALSE")%>%
  ggplot(., aes(hair_texture,
                      curv_median))+
  geom_violin(fill="#a6bddb",
              scale = "width")+
  geom_point(position = position_jitter(width=0.1),
             alpha = 0.4)+
  theme(axis.text.x = element_text(angle=90))+
  labs(x = "Self reported hair texture",
       y = bquote("Median curvature ("*mm^-1*")"),
       title = "Quantitative variation underlying \n self reported hair texture (median)")+
  theme_bw() +
  coord_flip()

plt_SelfRepCurv <- plot_grid(plt_admixed_SelfRepHair_CurvMean,
          plt_admixed_SelfRepHair_CurvMedian, labels = c("a", "b"))

plt_SelfRepCurv

```

The single individual with a high mean curvature in the straight group is the result of an artefact in the image. 

```{r artefact_im, fig.cap= "Image of hair sample with artefact biasing the measurement"}
# knitr for PDF document
# knitr::include_graphics("../docs/assets/artefact.jpg", error = FALSE)
# knitr for workflowr HTML
knitr::include_graphics("assets/artefact.jpg", error = FALSE)
```

The red arrow points to a stray fiber that likely contaminated the sample and was missed during imaging. Such potential outliers are the reason we chose to use the median curvature for a sample in our analyses.

# Objective hair texture classification vs. quantitative curvature

To explore how much data is lost when binning continuous variation, we compared mean and median curvature to classified hair texture. This classification is based on Loussouarn et al.'s 2007 paper ["Worldwide diversity of hair curliness: a new method of assessment](https://doi.org/10.1111/j.1365-4632.2007.03453.x). 

While the authors propose a number of parameters to distinguish curlier hair types (based on number of twists and waves among other factors), their primary classification is based on curvature. We demonstrate that, regardless of additional parameters, a considerable range of curvature is obscured when collapsing hair variation according to their curvature thresholds.

```{r plt_admixed_HairTypeBin, fig.cap= "Objective hair classification vs. quantitative curvature"}

plt_admixed_HairTypeBin_CurvMean <- df_admixed_plt %>%
  mutate(HairTypeBin_mean = 
           factor(case_when(
             curv_mean<= (1/54) ~ 'Straight',
             curv_mean<= (1/28.5) ~ 'Wavy',
             curv_mean<= (1/6) ~ 'Curly',
             TRUE ~ 'Very Curly'), 
             levels = c("Straight", "Wavy", "Curly", "Very Curly"))) %>% 
  ggplot(., aes(HairTypeBin_mean,
                      curv_mean))+
  geom_violin(fill="#a6bddb", scale = "width")+
  geom_point(position = position_jitter(width=0.1),
             alpha = 0.4)+
  theme(axis.text.x = element_text(angle=90))+
  labs(x = "Classified hair texture \n(based on median curvature)",
       y = bquote("Mean curvature ("*mm^-1*")")) +
  theme_bw() +
  coord_flip()


plt_admixed_HairTypeBin_CurvMedian <- df_admixed_plt %>%
  mutate(HairTypeBin_median = 
           factor(case_when(
             curv_median<= (1/54) ~ 'Straight',
             curv_median<= (1/28.5) ~ 'Wavy',
             curv_median<= (1/6) ~ 'Curly',
             TRUE ~ 'Very Curly'), levels = c("Straight", "Wavy", "Curly", "Very Curly"))) %>% 
  ggplot(., aes(HairTypeBin_median,
                      curv_median))+
  geom_violin(fill="#a6bddb", scale = "width")+
  geom_point(position = position_jitter(width=0.1),
             alpha = 0.4)+
  theme(axis.text.x = element_text(angle=90))+
  labs(x = "Classified hair texture \n(based on median curvature)",
       y = bquote("Median curvature ("*mm^-1*")"))+
  theme_bw() +
  coord_flip()

plt_HairBin <- plot_grid(plt_admixed_HairTypeBin_CurvMean,
          plt_admixed_HairTypeBin_CurvMedian, labels = c("a", "b"))


plt_HairBin
```


```{r imsave_QualQuant, include=FALSE}
# Saving different images for manuscript

plt_QualQuant_CurvMedian <- plot_grid((plt_admixed_SelfRepHair_CurvMedian + ggtitle(" Median curvature vs. \nSelf-reported hair texture")), (plt_admixed_HairTypeBin_CurvMedian + ggtitle("Median curvature vs. \nClassified hair texture")), labels = c("a", "b"))
  

pltsave_func(plt_QualQuant_CurvMedian, plot_path, width = 8, height = 4)

```

# Ancestry vs. hair morphology 

We carried out a number of analyses using the genotype data collected for this diverse sample. In an admixed sample where a continuous trait has divergent distributions in the parental ancestry groups, the resulting admixed population can show a correlation between ancestry and that trait. Finding such a correlation suggests may imply a polygenic trait with high heritability.

## Admixture components
Our sample consists of admixed individuals with primarily African and European ancestry.

```{r plt_admixture, fig.height=5, fig.width=10, fig.cap= "Admixture components for sample"}

anc <- df_admixed_plt %>% 
  ungroup() %>% 
  select(ID, SAS, AMR, AFR, EUR, EAS) %>% 
  arrange(AFR)

anc$ID <- factor(anc$ID, levels = anc$ID)

manc = reshape2::melt(anc, id.vars="ID")

plt_manc = ggplot(manc)+
  geom_bar(aes(ID,value, fill=variable),
           stat = "identity",
           width=1)+
  theme_bw()+
  theme(axis.text.x = element_blank(),
        axis.ticks.x = element_blank())+
  labs(x = "Individuals",
       y = "Ancestry",
       fill = "Population")+
  scale_fill_manual(values = c(
    "#e41a1c",
    "#377eb8",
    "#4daf4a",
    "#984ea3",
    "#ff7f00"
  ))

plt_manc

```

The colors represent ancestries that correspond to the following 1000 Genomes populations:
- SAS = South Asian
- AMR = American
- AFR = African
- EUR = European
- EAS = East Asian

Each of these are metapopulations based on the grouping of multiple (sub)continental population groups in the [1000 Genomes repository](https://www.internationalgenome.org/category/population/).

## Ancestry vs. curvature

Here we plot the correlation between proportion of African ancestry and m-index, median curvature, and eccentricity.

```{r stats_anc_corr, include=FALSE}
r2_mindex_afr = with(df_admixed_plt, cor(AFR, m_index,
                                         use = "complete.obs"))

r2_mindex_afr

r2_curv_afr = with(df_admixed_plt, cor(AFR, curv_median))

r2_curv_afr

r2_ecc_afr = with(df_admixed_plt, cor(AFR, eccentricity_median, 
                                      use = "complete.obs"))

r2_ecc_afr
```


```{r plt_anc_corr, fig.height=10, fig.width=7, fig.cap= "Percentage of African ancestry vs. M-index, curvature and eccentricity"}
plt_mindex_afr = ggplot(df_admixed_plt, 
              aes(AFR, m_index))+
  geom_point(alpha = 0.7)+
  stat_smooth(method = "lm")+
  theme_classic()+
  labs(x = "African Ancestry",
       y = bquote("Melanin index"))+
  annotate(geom="text",x = 0.2, y = 70,
           label = bquote(r^2*" = 0.83"))


plt_curv_afr = ggplot(df_admixed_plt, 
              aes(AFR, curv_median))+
  geom_point(alpha = 0.7)+
  stat_smooth(method = "lm")+
  theme_classic()+
  labs(x = "African Ancestry",
       y = bquote("Median curvature ("*mm^-1*")"))+
  annotate(geom="text",x = 0.2, y = 0.7,
           label = bquote(r^2*" = 0.85"))


plt_ecc_afr = ggplot(df_admixed_plt, 
              aes(AFR, eccentricity_median))+
  geom_point(alpha = 0.7)+
  stat_smooth(method = "lm")+
  theme_classic()+
  labs(x = "African Ancestry",
       y = bquote("Eccentricity"))+
  annotate(geom="text",x = 0.3, y = 0.5,
           label = bquote(r^2*" = 0.56"))


plt_trait_afr <- plot_grid(plt_mindex_afr, plt_curv_afr, plt_ecc_afr, labels = c("a", "b", "c"), nrow = 3, ncol = 1)


plt_trait_afr

```


## Curvature vs. eccentricity

The relationship between cross-sectional shape (eccentricity) and curvature has long been debated. Due to the coincidence of cross-sectional shape and curvature in various populations that are often contrasted (i.e. East Asian vs. North European vs. West African), it has been unclear whether these traits have a causal relationship (specifically that higher eccentricity predicts higher curvature). In our admixed sample, we have the opportunity to test this question and fit a model between these traits with and without ancestry.

```{r df_curv_ecc}
df_curv_ecc = df_admixed_plt %>%
  ungroup() %>%
  select(curv_median, eccentricity_median, AFR)%>%
  drop_na()
```

### Uncorrected

First we examine the data without correcting for ancestry.

```{r plt_curv_ecc, fig.cap= "Curvature vs. eccentricity (without correction for ancestry)"}


evc1 = ggplot(df_curv_ecc, aes(eccentricity_median,
                               curv_median))+
  geom_point()+
  theme_classic()+
  stat_smooth(method="lm")+
  labs(y = "Curvature",
       x = "Eccentricity (median)",
       title = "Curvature vs eccentricity \n (ancestry uncorrected)")+
  annotate(geom = "text",
          x = 0.4, y = 0.4, hjust = 0,
          label = "p-value = 1.96e-08")+
  ylim(c(0.01,0.8))

evc1

```

Fit a simple linear model and check the residuals. 

```{r stats_curv_ecc}
lcurv_ecc.1 = lm(data = df_curv_ecc, 
                 curv_median ~ eccentricity_median)


summary(lcurv_ecc.1)

#write function to plot residuals

plot(lcurv_ecc.1, which = 1)

```

The residual plots show that there might be a few problems with the relationship:
1. possibly an outlier distorting the relationship, 
2. relationship might not be linear
3. errors might be heteroescedastic.

Let's refit the model after removing the outlier. 

```{r}

df_curv_ecc = df_curv_ecc %>%
  filter(eccentricity_median > 0.4)

lcurv_ecc.1 = lm(data = df_curv_ecc, 
                 curv_median ~ eccentricity_median)


summary(lcurv_ecc.1)

plot(lcurv_ecc.1, which = 1)

```

The outlier removal improves the fit. Next, check if the data fits a curvilinear (quadratic) model better.

```{r}
#center eccentricity first 
df_curv_ecc = df_curv_ecc%>%
  mutate(eccentricity_median.c = scale(eccentricity_median, scale = FALSE),
    eccentricity_median.c2 = eccentricity_median.c^2)


lcurv_ecc.1.1 = lm(data = df_curv_ecc, 
                 curv_median ~ eccentricity_median.c + eccentricity_median.c2)

plot(lcurv_ecc.1.1, which = 1)

```

The quadratic fit is better that the simple linear fit. Let's compare the two models formally and see if the quadratic fit is better.

```{r}

anova(lcurv_ecc.1, lcurv_ecc.1.1, test = "Chisq")

```

This shows that the residual sum of squares is significantly smaller for the quadratic fit, i.e., the quadratic model fits the data better.

There still seems to be some heteroescedasticity in the errors. We will use the fitted values as weights in a weighted regression. Because the variance increases with the fitted values, we will use $\frac{1}{fitted ~ values^2}$ as weights. 

```{r}

#weights
df_curv_ecc$inv.curv = 1/lcurv_ecc.1.1$fitted.values^2
  
lcurv_ecc.1.2 = lm(data = df_curv_ecc, 
                 curv_median ~ eccentricity_median.c +
                   eccentricity_median.c2, 
                 weights = inv.curv)

summary(lcurv_ecc.1.2)

#for this, we need to plot the standardized residuals
plot(fitted(lcurv_ecc.1.2),
     rstandard(lcurv_ecc.1.2),
     xlab = "Fitted values\n lm(curv_median ~ eccentricity_median + eccentricity_median^2)",
     ylab = "Standardized residuals", 
     main = "Residuals vs fitted\n(weighted regression)")

```

The variance in the residuals is more homogeneous under this model. We will re-generate the scatterplot with this model fit.

```{r}

df_curv_ecc$fitted.cve = lcurv_ecc.1.2$fitted.values

# r2_cve = paste("r^2 ~", round(summary(lcurv_ecc.1.2)$r.squared, 2))

pvalue_cve1 = formatC(summary(lcurv_ecc.1.2)$coefficients[2,4],2)
pvalue_cve2 = formatC(summary(lcurv_ecc.1.2)$coefficients[3,4],2)


plt_cve = ggplot(df_curv_ecc, 
       aes(eccentricity_median,
                               curv_median)) +
  geom_point()+
  theme_classic()+
  labs(y = "Curvature",
       x = "Eccentricity (median)",
       title = "Curvature vs eccentricity \n (ancestry uncorrected)")+
  annotate(geom = "text",
          x = 0.5, y = 0.7, hjust = 0,
          label = paste("pvalue (eccentricity):", 
                        pvalue_cve1))+
    annotate(geom = "text",
          x = 0.5, y = 0.6, hjust = 0,
          label = paste("pvalue (eccentricity^2):", 
                        pvalue_cve2))+
  ylim(c(0.01,0.8)) +
  geom_line(aes(eccentricity_median, fitted.cve), 
             color = "red", size = 1)

plt_cve

```

Thus, there is a significant relationship between eccentricity and curvature. Does this hold when we add ancestry as a covariate in the model?

### Corrected

We re-plot the relationship after residualizing curvature and eccentriciy (both) on ancestry. 

```{r plt_curv_ecc_anc, fig.cap= "Curvature vs. eccentricity (with correction for ancestry)"}


lcurv_afr = lm(data = df_curv_ecc, 
                 curv_median ~ AFR)

lecc_afr = lm(data = df_curv_ecc, 
                 eccentricity_median ~ AFR)


#turns out the correlation between curvature and eccentricity is due to stratification.
df_curv_ecc = df_curv_ecc %>%
  mutate(resid_cvafr = resid(lcurv_afr),
         resid_evafr = resid(lecc_afr))

evc2 = ggplot(df_curv_ecc, 
              aes(resid_evafr,
                  resid_cvafr))+
  geom_point()+
  theme_classic()+
  stat_smooth(method="lm")+
  labs(y = "Curvature (ancestry corrected)",
       x = "Eccentricity (ancestry corrected)",
       title = "Curvature vs eccentricity \n (ancestry corrected)")
  
evc2

```

When we correct for ancestry, the association between eccentricity and curvature is no longer significant. This supports the idea that these traits co-occur due to ancestry stratification. Let's test this formally.

The relationship looks rather linear so we will start with a simple linear model and inspect the residuals. 

```{r}

lcve_corrected = lm(data = df_curv_ecc, 
                    curv_median ~ eccentricity_median + AFR)

summary(lcve_corrected)

plot(lcve_corrected, which = 1)

```

The relationship looks pretty linear but there's still some heteroescedasticity in the residuals. We will again fit a weighted regression model.

```{r}

df_curv_ecc$fitted.cve_afr = 1/fitted(lcve_corrected)^2

lcve_corrected.1 = lm(data = df_curv_ecc, 
                    curv_median ~ eccentricity_median + AFR, weights = fitted.cve_afr)

summary(lcve_corrected.1)

plot(lcve_corrected.1$fitted.values, rstandard(lcve_corrected.1), 
     xlab = "Fitted values\n lm(curv_median ~ eccentricity_median + AFR)", 
     ylab = "Standardized residuals",
     main = "Residuals vs fitted values\n(Weighted regression model)")

```

The weighted regression accounts for the heteroescedasticity.

```{r}

df_curv_ecc$fitted.cve_afr.weighted = fitted(lcve_corrected.1)

pvalue_cve.corrected = round(summary(lcve_corrected.1)$coefficients[2,4],2)

evc2.1 = ggplot(df_curv_ecc, 
              aes(resid_evafr,
                  resid_cvafr))+
  geom_point()+
  theme_classic()+
  labs(y = "Curvature (ancestry corrected)",
       x = "Eccentricity (ancestry corrected)",
       title = "Curvature vs eccentricity \n (ancestry corrected)")+
  stat_smooth(method = "lm", se = FALSE, color = "red", size = 1)+
  annotate(geom = "text",
          x = -0.25, y = 0.2, hjust = 0,
          label = paste("pvalue (eccentricity):", pvalue_cve.corrected))

evc2.1

```


## Curvature vs. skin pigmentation

To demonstrate the potential effect of population stratification on traits, we compare hair curvature with skin pigmentation (m-index). These two traits are not biologically related, yet, in an admixed population, we may see a correlation that is due to population stratification of these polygenic traits.

### Uncorrected

First we examine the relationship between curvature and skin pigmentation without correcting for ancestry.

```{r stats_curv_mindex}

df_curv_mindex = df_admixed_plt %>%
  ungroup() %>%
  select(curv_median, m_index, AFR)%>%
  drop_na()

lcurv_mindex.1 = lm(data = df_curv_mindex, 
                    curv_median ~ m_index)

summary(lcurv_mindex.1)

plot(lcurv_mindex.1, which = 1)

```


```{r plt_curv_mindex, fig.cap= "Curvature vs. M-index (without correction for ancestry)"}
mvc_uncorrected = ggplot(df_curv_mindex,
                         aes(m_index,
                                    curv_median))+
  geom_point()+
  theme_classic()+
  labs(x = "Melanin index",
       y = "Curvature",
       title = "Curvature vs melanin index \n (ancestry uncorrected)")+
  stat_smooth(method="lm")+
  annotate(geom = "text",
           x = 50, y = 0.15, 
           label = "p-value < 2e-16")

mvc_uncorrected

```

As expected, we see a significant correlation between the two traits. The relationship appears to be linear but there seems to be more variance for observations with higher melanin index/curvature. We will use a weighted regression approach again.

```{r}

df_curv_mindex$weights_cvm = 1/lcurv_mindex.1$fitted.values^2

lcurv_mindex.2 = lm(data = df_curv_mindex, 
                    curv_median ~ m_index, 
                    weights = weights_cvm)

summary(lcurv_mindex.2)

plot(fitted(lcurv_mindex.2), 
     rstandard(lcurv_mindex.2),
     xlab = "Fitted values",
     ylab = "Standardized residuals")

```

That fixes the heteroescedasticity issue.

```{r}

pvalue_cvm.weighted = formatC(summary(lcurv_mindex.2)$coefficients[2,4])

mvc_uncorrected = ggplot(df_curv_mindex,
                         aes(m_index,
                                    curv_median))+
  geom_point()+
  theme_classic()+
  labs(x = "Melanin index",
       y = "Curvature",
       title = "Curvature vs melanin index \n (ancestry uncorrected)")+
  stat_smooth(method="lm", color = "red", se = FALSE)+
    annotate(geom = "text",
           x = 50, y = 0.15, hjust = 0,
           label = paste("pvalue (MIndex):\n", 
           pvalue_cvm.weighted))

mvc_uncorrected

```


### Corrected

We then apply a correction for ancestry and re-analyze the data. 

```{r stats_curv_mindex_anc}

lcurv_afr.2 = lm(data = df_curv_mindex, 
                    curv_median ~ AFR)

lmindex_afr.1 = lm(data = df_curv_mindex, 
                 m_index ~ AFR)

df_curv_mindex = df_curv_mindex %>%
  mutate(resid_cvafr = rstandard(lcurv_afr.2), 
         resid_miafr = rstandard(lmindex_afr.1))

lmindex_curv = lm(data = df_curv_mindex, 
                  curv_median ~ m_index + AFR, 
                  weights = weights_cvm)

pvalue_cvm.corrected = round(summary(lmindex_curv)$coefficients[2,4],3)

mvc_corrected = ggplot(df_curv_mindex,
                       aes(resid_miafr, 
                           resid_cvafr))+
  geom_point()+
  theme_classic()+
  labs(x = "Melanin index (ancestry corrected)",
       y = "Curvature (ancestry corrected)",
       title = "Curvature vs melanin index \n(ancestry corrected)")+
  stat_smooth(method="lm",se = FALSE, color = "red")+
    annotate(geom = "text",
           x = 0, y = -3, 
           label = paste("pvalue (MIndex):", pvalue_cvm.corrected))

mvc_corrected

```

Like with curvature and eccentricity, the relationship between curvature and skin pigmentation is no longer significant when ancestry is taken into account. 

```{r imsave_mvc_evc_combo, include=FALSE}

plt_strat <- plot_grid(mvc_uncorrected, mvc_corrected, plt_cve, evc2.1, labels = c("a", "b", "c", "d"), nrow = 2, ncol = 2)

pltsave_func(plt_strat, plot_path, width = 8, height = 8)
```